File:FS RC dia.png

Shows the largest circle within a right isosceles triangle.

Base is the right isosceles triangle of side length ${\displaystyle a}$ and centroid ${\displaystyle S_{0}}$

Inscribed is the largest possible circle with radius ${\displaystyle r_{1}}$ around point ${\displaystyle S_{1}}$

0) The side length of the base right triangle ${\displaystyle |AB|=|AC|=a}$
1) Radius of the circle ${\displaystyle r_{1}={\frac {2-{\sqrt {2}}}{2}}\cdot a\quad }$ (See calculation 1).

0) Perimeter of base triangle ${\displaystyle P_{0}=|AB|+|BC|+|AC|=a+{\sqrt {2}}\cdot a+a=(2+{\sqrt {2}})\cdot a}$
1) Perimeter of the circle ${\displaystyle P_{1}=\pi \cdot (2-{\sqrt {2}})\cdot a\quad }$ (See calculation 2 )

0) Area of the base triangle ${\displaystyle A_{0}={\frac {a^{2}}{2}}}$
1) Area of the inscribed circle ${\displaystyle A_{1}=\pi \left({\frac {3-2{\sqrt {2}}}{2}}\right)\cdot a^{2}\quad }$ (See calculation 3)

Centroid positions are measured from the centroid point of the base shape
0) Centroid positions of the base triangle: ${\displaystyle x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed circle: ${\displaystyle x_{1}=y_{1}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a\quad }$, (see calculation 4)

In the normalised case the area of the base is set to 1.
${\displaystyle A_{0}=1\quad \Rightarrow {\frac {a^{2}}{2}}=1\quad \Rightarrow a={\sqrt {2}}}$

0) Side length of the base triangle ${\displaystyle a={\sqrt {2}}\quad \approx 1.414}$
1) Radius of the inscribed circle ${\displaystyle r_{1}={\frac {2-{\sqrt {2}}}{2}}\cdot {\sqrt {2}}={\frac {2{\sqrt {2}}-2}{2}}={\sqrt {2}}-1\quad \approx 0.414}$

0) Perimeter of base triangle ${\displaystyle P_{0}=(2+{\sqrt {2}})\cdot {\sqrt {2}}=2{\sqrt {2}}+2\quad \approx 4.8284271}$

1) Perimeter of the inscribed circle ${\displaystyle P_{1}=\pi \cdot (2-{\sqrt {2}})\cdot {\sqrt {2}}=\pi \cdot (2{\sqrt {2}}-2)\quad \approx 2.6025805}$
S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}\quad \approx 7.431007}$

0) Area of the base triangle ${\displaystyle A_{0}=1}$
1) Area of the inscribed circle ${\displaystyle A_{1}=\pi \left({\frac {3-2{\sqrt {2}}}{2}}\right)\cdot \left({\sqrt {2}}\right)^{2}=\pi (3-2{\sqrt {2}})\quad \approx 0.539}$

Centroid positions are measured from the centroid point of the base shape.
0) Centroid positions of the base triangle: ${\displaystyle S_{0}:\quad x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed circle: ${\displaystyle S_{1}:x_{1}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot {\sqrt {2}}={\frac {4{\sqrt {2}}-6}{6}}={\frac {2{\sqrt {2}}-3}{3}}=-0.057...\quad y_{1}=x_{1}==-0.057...}$

The distance between the centroid of the base element and the centroid of the circle is:
${\displaystyle d={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.0808802...}$

Apart of the base element there is only one shape allocated. Therefore the integer part of the identifying number is 1.
The decimal part of the identifying number is the decimal part of the sum of the perimeters and the distances of the centroids in the normalised case.

${\displaystyle decimalpart(7.4310077...+0.0808802...)=decimalpart(7.5118879...)=.5118879}$

So the identifying number is: ${\displaystyle 1.5118879}$

Because FS_R is a right isosceles triangle the following equations hold:
(1) ${\displaystyle \quad |AB|=|AC|=a}$
(2) ${\displaystyle \quad |BC|={\sqrt {2}}\cdot a}$
(3) ${\displaystyle \quad |BD|=|CD|={\frac {{\sqrt {2}}\cdot a}{2}}={\frac {a}{\sqrt {2}}}}$
The vertices ${\displaystyle A,x_{1},S_{1},y_{1}}$ form a square with side length ${\displaystyle r_{1}}$. This leads to equation:
(4) ${\displaystyle \quad |AS_{1}|={\sqrt {2}}\cdot r_{1}}$
D is the tangent point of line segment ${\displaystyle [BC]}$ and the circle around ${\displaystyle S_{1}}$ with radius ${\displaystyle r_{1}}$. This gives:
(5) ${\displaystyle \quad |DS_{1}|=r_{1}}$.
Applying equations (4) and (5) gives:
(6) ${\displaystyle \quad |AD|=|AS_{1}|+|DS_{1}|={\sqrt {2}}\cdot r_{1}+r_{1}=({\sqrt {2}}+1)\cdot r_{1}}$.

The triangle ${\displaystyle \triangle ABD}$ is a right triangle where the Pythagorean theorem holds:
${\displaystyle \quad |AD|^{2}+|BD|^{2}=|AB|^{2}}$
${\displaystyle \quad \Leftrightarrow |AD|^{2}=|AB|^{2}-|BD|^{2}}$
${\displaystyle \quad \Leftrightarrow |AD|^{2}=|AB|^{2}-|BD|^{2}}$
${\displaystyle \quad \Leftrightarrow |AD|^{2}=a^{2}-|BD|^{2}\quad }$, using (1)
${\displaystyle \quad \Leftrightarrow |AD|^{2}=a^{2}-\left({\frac {a}{\sqrt {2}}}\right)^{2}\quad }$, using (3)
${\displaystyle \quad \Leftrightarrow |AD|^{2}=a^{2}-{\frac {a^{2}}{2}}\quad }$
${\displaystyle \quad \Leftrightarrow |AD|^{2}={\frac {a^{2}}{2}}\quad }$

${\displaystyle \quad \Leftrightarrow |AD|={\frac {a}{\sqrt {2}}}\quad }$
${\displaystyle \quad \Leftrightarrow ({\sqrt {2}}+1)\cdot r_{1}={\frac {a}{\sqrt {2}}}\quad }$, using (6)
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {a}{({\sqrt {2}}+1)\cdot {\sqrt {2}}}}\quad }$
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {a}{(2+{\sqrt {2}})}}\quad }$
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {a}{(2+{\sqrt {2}})}}\cdot {\frac {(2-{\sqrt {2}})}{(2-{\sqrt {2}})}}\quad }$
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {2-{\sqrt {2}}}{2}}\cdot a\quad }$

Perimeter of the circle:
${\displaystyle \quad P_{1}=2\pi \cdot r_{1}}$
${\displaystyle \quad \Leftrightarrow P_{1}=2\pi \cdot {\frac {2-{\sqrt {2}}}{2}}\cdot a}$
${\displaystyle \quad \Leftrightarrow P_{1}=\pi \cdot (2-{\sqrt {2}})\cdot a}$

Area of the inscribed circle:
${\displaystyle \quad A_{1}=\pi \cdot r_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}=\pi \left({\frac {2-{\sqrt {2}}}{2}}\cdot a\right)^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}=\pi \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}\cdot a^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}=\pi \left({\frac {4-2\cdot 2\cdot {\sqrt {2}}+{\sqrt {2}}{\sqrt {2}}}{4}}\right)\cdot a^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}=\pi \left({\frac {4-4{\sqrt {2}}+2}{4}}\right)\cdot a^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}=\pi \left({\frac {6-4{\sqrt {2}}}{4}}\right)\cdot a^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}=\pi \left({\frac {3-2{\sqrt {2}}}{2}}\right)\cdot a^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}=\pi (3-2{\sqrt {2}})\left({\frac {a^{2}}{2}}\right)}$
${\displaystyle \quad \Leftrightarrow A_{1}=\pi (3-2{\sqrt {2}})\cdot A_{0}}$

Centroid of the inscribed circle measured from the centroid of the base triangle:
${\displaystyle \quad x_{1}=|Ax_{1}|-|Ax_{0}|}$
${\displaystyle \quad \Leftrightarrow x_{1}=|Ax_{1}|-{\frac {1}{3}}\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{1}=|AS_{1}|\cdot cos{45}-{\frac {1}{3}}\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{1}=|AS_{1}|\cdot {\frac {1}{\sqrt {2}}}-{\frac {1}{3}}\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{1}={\sqrt {2}}\cdot r_{1}\cdot {\frac {1}{\sqrt {2}}}-{\frac {1}{3}}\cdot a\quad }$, using (4)
${\displaystyle \quad \Leftrightarrow x_{1}=r_{1}-{\frac {1}{3}}\cdot a\quad }$, using (4)
${\displaystyle \quad \Leftrightarrow x_{1}={\frac {2-{\sqrt {2}}}{2}}\cdot a-{\frac {1}{3}}\cdot a\quad }$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {2-{\sqrt {2}}}{2}}-{\frac {1}{3}}\right)\cdot a\quad }$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {6-3{\sqrt {2}}}{6}}-{\frac {2}{6}}\right)\cdot a\quad }$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a\quad }$

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